Sure, here is a Python solution for the problem:

```python
def palindrome_pairs(words):
def is_palindrome(check):
return check == check[::-1]

words = {word: i for i, word in enumerate(words)}
valid_pals = []
for word, k in words.items():
n = len(word)
for j in range(n+1):
pref = word[:j]
suf = word[j:]
if is_palindrome(pref):
back = suf[::-1]
if back != word and back in words:
valid_pals.append((words[back], k))
if j != n and is_palindrome(suf):
back = pref[::-1]
if back != word and back in words:
valid_pals.append((k, words[back]))
return valid_pals

print(palindrome_pairs(["code", "edoc", "da", "d"]))
```

This solution works by:

1. Creating a dictionary where the key is a word from the list and the value is the index of that word.
2. Iterating over each word in the dictionary, and for each word, it checks every possible prefix and suffix of the word.
3. If the prefix is a palindrome, it reverses the suffix and checks if this reversed suffix exists in the dictionary (and is not the word itself). If it does, it adds the pair to the result list.
4. If the suffix is a palindrome and is not the whole word (to avoid duplicating the result), it reverses the prefix and checks if this reversed prefix exists in the dictionary. If it does, it adds the pair to the result list.
5. Finally, it returns the list of valid pairs.

Question

Sure, here is a Python solution for the problem:

```python
def palindrome_pairs(words):
    def is_palindrome(check):
        return check == check[::-1]

words = {word: i for i, word in enumerate(words)}
    valid_pals = []
    for word, k in words.items():
        n = len(word)
        for j in range(n+1):
            pref = word[:j]
            suf = word[j:]
            if is_palindrome(pref):
                back = suf[::-1]
                if back != word and back in words:
                    valid_pals.append((words[back], k))
            if j != n and is_palindrome(suf):
                back = pref[::-1]
                if back != word and back in words:
                    valid_pals.append((k, words[back]))
    return valid_pals

print(palindrome_pairs(["code", "edoc", "da", "d"]))
```

This solution works by:

1. Creating a dictionary where the key is a word from the list and the value is the index of that word.
2. Iterating over each word in the dictionary, and for each word, it checks every possible prefix and suffix of the word.
3. If the prefix is a palindrome, it reverses the suffix and checks if this reversed suffix exists in the dictionary (and is not the word itself). If it does, it adds the pair to the result list.
4. If the suffix is a palindrome and is not the whole word (to avoid duplicating the result), it reverses the prefix and checks if this reversed prefix exists in the dictionary. If it does, it adds the pair to the result list.
5. Finally, it returns the list of valid pairs.

Knowee AI · Accepted Answer

Sure, here is a Python solution for the problem:

```python
def palindrome_pairs(words):
    def is_palindrome(check):
        return check == check[::-1]

words = {word: i for i, word in enumerate(words)}
    valid_pals = []
    for word, k in words.items():
        n = len(word)
        for j in range(n+1):
            pref = word[:j]
            suf = word[j:]
            if is_palindrome(pref):
                back = suf[::-1]
                if back != word and back in words:
                    valid_pals.append((words[back], k))
            if j != n and is_palindrome(suf):
                back = pref[::-1]
                if back != word and back in words:
                    valid_pals.append((k, words[back]))
    return valid_pals

print(palindrome_pairs(["code", "edoc", "da", "d"]))
```

This solution works by:

1. Creating a dictionary where the key is a word from the list and the value is the index of that word.
2. Iterating over each word in the dictionary, and for each word, it checks every possible prefix and suffix of the word.
3. If the prefix is a palindrome, it reverses the suffix and checks if this reversed suffix exists in the dictionary (and is not the word itself). If it does, it adds the pair to the result list.
4. If the suffix is a palindrome and is not the whole word (to avoid duplicating the result), it reverses the prefix and checks if this reversed prefix exists in the dictionary. If it does, it adds the pair to the result list.
5. Finally, it returns the list of valid pairs.

concatenation of the two wordsGiven a list of words, find all pairs of unique indices such that the concatenation of the two words is a palindrome.For example,given the list ["code", "edoc", "da", "d"], return [(0, 1), (1, 0), (2, 3)].

Question

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